1ſhall be the Time of the whole A F, and E C the Time of A I: And be
cauſe in the Equicrural Triangle A E D the Side A E is equal to the
Side E D, E D ſhall be the Time along A F, and E C is the Time along
A I: Therefore C D, that is A B ſhall be the Time along A F ex qui
ete in A; which is the ſame as if we ſaid, that A B is the Time along
A G out of G, or out of H: Which was to be done.
cauſe in the Equicrural Triangle A E D the Side A E is equal to the
Side E D, E D ſhall be the Time along A F, and E C is the Time along
A I: Therefore C D, that is A B ſhall be the Time along A F ex qui
ete in A; which is the ſame as if we ſaid, that A B is the Time along
A G out of G, or out of H: Which was to be done.
PROBL. XIII. PROP. XXXIV.
An inclined Plane and Perpendicular whoſe ſub
lime term is the ſame being given, to find a
more ſublime point in the Perpendicular pro
longed out of which a Moveable falling, and
being turned along the inclined Plane, may
paſſe them both in the ſame Time, as it doth
the ſole inclined Plane ex quiete in its ſuperi
our Term.
lime term is the ſame being given, to find a
more ſublime point in the Perpendicular pro
longed out of which a Moveable falling, and
being turned along the inclined Plane, may
paſſe them both in the ſame Time, as it doth
the ſole inclined Plane ex quiete in its ſuperi
our Term.
Let the inclined Plane and Perpendicular be A B and A C, whoſe
Term A is the ſame. It is required in the Perpendicular prolonged
from A to find a ſublime point, out of which the Moveable deſcen
ding, and being turned along the Plane A B, may paſſe the aſſigned part
of the Perpendicular and the Plane A B in the ſame Time, as it would the
ſole Plane A B out of Reſt in A.
Term A is the ſame. It is required in the Perpendicular prolonged
from A to find a ſublime point, out of which the Moveable deſcen
ding, and being turned along the Plane A B, may paſſe the aſſigned part
of the Perpendicular and the Plane A B in the ſame Time, as it would the
ſole Plane A B out of Reſt in A.
133[Figure 133]Let the Ho
rizontal Line
be B C, and
let A N be
cut equal to
A C; and as
A B is to B N,
ſo let A L be
to L C: and
unto A L let
A I be equal,
and unto A C
and B I let C
E be a third
proportional,
marked in the
Perpendicular A C produced. I ſay, that C E is the Space acquired;
ſo that the Perpendicular being extended above A, and the part A X
equal to C E being taken, a Moveable out of X will paſſe both the
rizontal Line
be B C, and
let A N be
cut equal to
A C; and as
A B is to B N,
ſo let A L be
to L C: and
unto A L let
A I be equal,
and unto A C
and B I let C
E be a third
proportional,
marked in the
Perpendicular A C produced. I ſay, that C E is the Space acquired;
ſo that the Perpendicular being extended above A, and the part A X
equal to C E being taken, a Moveable out of X will paſſe both the